A refrigerator converts 100 g of water at 20° C to ice at -10° C in 73.5 min. Calculate the average rate of heat extraction in watt. The specific heat capacity of water is 4.2 J g^-1 K^-1 , specific latent heat of ice is 336 J g^-1 and the specific heat capacity of ice is 2.1 J g^-1 K^-1.

Given,

mass (m) = 100 g

time (t) = 73.5 min

specific heat capacity of water = 4.2 J g^-1 K^-1

specific latent heat of ice = 336 J g^-1

specific heat capacity of ice = 2.1 J g^-1 K^-1

Heat energy released by water in fall of it's temperature from 20° to 0° C (Q₁)
= mass x specific heat capacity x fall in temperature
= 100 × 4.2 x (20 - 0)
= 100 × 4.2 x 20
= 8400 J

Hence, Q₁ = 8400 J

Heat energy released by water when it converts into ice at 0° C (Q₂) = m x L_ice
= 100 × 336
= 33600 J

Heat energy released when ice cools from 0° C to -10° C (Q₃) = m x c x change in temperature
= 100 × 2.1 x [0 - (-10)]
= 100 x 2.1 x 10
= 2100 J

Hence,

Total heat energy = Q₁ + Q₂ + Q₃
= 8400 + 33600 + 2100
= 44100 J

Time taken = 73.5 min = 4410 s

Average rate of heat extraction (P)

$P = \dfrac{E}{t}$

Substituting the values in the formula we get,

$P = \dfrac{44100}{4410} \\[0.5em] P = 10 W$

Hence, average rate of heat extraction = 10 W